Uranium said:
I've been reading a bit about hybrid methods, and I keep coming across adjoint fluxes. What exactly is an adjoint flux? And how does it factor into calculations?
Uranium,
Mathematical operators have "adjoints". For example, the adjoint of a real matrix is its
transpose. The adjoint of the derivative is its negative.
You can take the Boltzmann transport equation, and take the adjoint of each of its
terms, and you get the adjoint Boltzmann equation. The solution to this equation is
the adjoint flux.
There is a physical interpretation to the adjoint equation; the concept of neutron
"importance" obeys the adjoint equation.
When you define certain quantities like reactivity, you do so with a weighting function
in general. If you use the adjoint flux as your weighting function, then your calculation
of the quantity becomes accurate to second order instead of first order when you use
first order perturbation theory.
Let H be the Boltzmann transport equation operator, and H* is its adjoint.
We have then psi as the solution to the forward equation - H psi = 0
We also have (H*)(psi*) = 0.
Suppose H' = H + dH and psi' = psi + dpsi
<psi*| H' | psi'> = <psi*| H + dH | psi + dpsi > =
<psi*| H | psi > + <psi*| H | dpsi> + <psi*| dH | psi > + <psi*| dH | dpsi >
The first term is zero [ H psi = 0 ]. First order perturbation theory would have one
using the third term as the answer. But the second term which we can't evaluate without
solving the new system for dpsi is also first order. However, from the definition of adjoints;
<psi*|H|dpsi> = < psi* | H dpsi > = < H*psi*|dpsi> = 0 since H* psi* = 0
Therefore, if you use only the 3rd term as per first order perturbation theory - your
error - the 4 th term is 2nd order - it has a dH and a dpsi.
The other first order term vanishes if you weight with the adjoint.
Dr. Gregory Greenman